Let \(f\) be analytic on the unit disk. Is it possible that \(f\bigl(\frac{1}{n}\bigr)\) takes the following values for \(n=2,3,4,\ldots\)? Why or why not?
\((-1)^n\)
\(e^{-n}\) for \(n\) even, and \(0\) for \(n\) odd
\(\dfrac{1}{\lfloor\sqrt{n}\rfloor}\) (here \(\lfloor x\rfloor\) is the greatest integer not greater than \(x\))
\(\dfrac{n-2}{n-1}\)
Solution
Since \(1/n\to0\), continuity and the identity theorem impose strong constraints.
(a) No. The values \((-1)^n\) do not converge as \(n\to\infty\), whereas continuity at \(0\) would force \[ f(1/n)\longrightarrow f(0). \]
(b) No. For every odd \(n\) we have \(f(1/n)=0\), and those zeros accumulate at the interior point \(0\). By the identity theorem, \(f\equiv0\), contradicting \(f(1/n)=e^{-n}\ne0\) for even \(n\).
(c) No. The prescribed values tend to \(0\), so necessarily \(f(0)=0\). If \(f\) were not identically zero, let \(m\ge1\) be the order of its zero at \(0\). Then \[ f(z)=z^m g(z),\qquad g(0)\ne0, \] so \[ n^m f(1/n)=g(1/n)\longrightarrow g(0). \] But the prescribed values give \[ n^m f(1/n)=\frac{n^m}{\lfloor\sqrt n\rfloor}\longrightarrow\infty, \] a contradiction.
(d) Yes. The function \[ f(z)=\frac{1-2z}{1-z} \] is holomorphic on the unit disk and satisfies \[ f(1/n)=\frac{1-2/n}{1-1/n}=\frac{n-2}{n-1}. \]